A carpet of mass `M` is rolled along its length so as to from a cylinder of radius `R` and is kept on a rough floor. When a negligibly small push is given to the cylindrical carpet, it stars rolling itself without sliding on the floor. Calculate horizontal velocity of cylindrical part of the carpet when its radius reduces to `R//2`.

Explanation

Figure shows the forces acting on the carpet in two cases:
i. when the radius of the roll is `R`
ii when the radius is `R/2`
Mass of the roll of radius `R/2=M/4`
(Mass is proportional to the carpe spread on the floor `=(3/4)M`. There has to be a frictional force that is too static in nature on the carpet once it starts unrolling . otherwise, why should the centre of mass of the system (carpet of mass `m`) move towards the right? Is such an unrolling possible on a smooth surface? Does the centre of mass of the system (`M`) move in the vertical direction? If yes, why?

Here `W_(f_(s))=0, W_(N)=0`
Therefore `W_("ncf")=0`
The mechanical energy of the system will reamin constant. The part of the carpet spread on the ground will have neither kinetic energy nor potential energy.
Therefore `MgR=(M/4)g(R/2)+1/2(M/4)v_(CM)^(2)+1/2(1/2M/4(R/2)2)omega^(2)`
`implies 7/8 MgR=M/8v_(CM)^(2) +1/16M(R/2)^(2)((v_(CM))/(R/2)^(2))`
`(v_(CM)=R/2omega)`
`7/8MgR=3/8mv_(CM)^(2)impliesv_(CM)=sqrt(14/3gR)`